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Physics - SHM Question with Solution | TestHub

PhysicsSHMEnergyEasy2 minPYQ_2020

PhysicsEasysingle choice

A particle starts oscillating simple harmonically from its equilibrium position with time period $T$ . The ratio of KE and PE of the particle at the $t = T / 12$ is

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Answer:

Solution:

When $t = \frac{T}{12}, t h e n x = A \sin \frac{2 π}{T} \times \frac{T}{12} = \frac{A}{2}$

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$

$= \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{4})$

$= \frac{3}{4} (\frac{1}{2} m ω^{2} A^{2})$

$P E = \frac{1}{2} m ω^{2} x^{2} = \frac{1}{4} (\frac{1}{2} m ω^{2} A^{2})$

$\frac{K E}{P E} = \frac{3}{1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: PYQ_2020

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